$56z^3-70z^2-21z-4=0$ how to solve for $z$. I formed three equations but not getting the answer. If I get a start or suggestion it would be a great help.
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1Is this by any chance motivated by this question? – JimmyK4542 Dec 22 '15 at 09:20
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1The rational root theorem does not yield any solutions to this. So it will most likely consist of 3 real solutions, or 1 real and 2 complex solutions. Look into the cubic formula, it will give you your solutions. – ultrainstinct Dec 22 '15 at 09:21
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The question is almost similar – Archis Welankar Dec 22 '15 at 09:31
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is $z$ a real variable? – Dr. Sonnhard Graubner Dec 22 '15 at 10:10
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I dont have solutions all i want is solutions complex or real – Archis Welankar Dec 22 '15 at 10:23
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Did you try Cardano ? – Claude Leibovici Dec 22 '15 at 10:29
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I saw it but it would be good if someone explained a bit more rather than just giving the site names – Archis Welankar Dec 22 '15 at 10:52
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I am not suggesting you to go to the site but to learn the method from the site. You must know how the method works. – Claude Leibovici Dec 22 '15 at 11:16
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Sorry if i have been rude but im in 11th std to understand so that is way to far so i asked it here – Archis Welankar Dec 22 '15 at 11:22
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1I think. You should try synthetic division. – Syed Muhammad Asad Jan 11 '16 at 15:03
2 Answers
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HINT:
$$56z^3-70z^2-21z-4=0\Longleftrightarrow$$
Eliminate the quadratic term by substituting $x=z-\frac{5}{12}$:
$$56x^3-\frac{301x}{6}-\frac{563}{27}=0\Longleftrightarrow$$ $$x^3-\frac{43x}{48}-\frac{563}{1512}=0\Longleftrightarrow$$
If $x=y+\frac{\lambda}{y}$ then $y=\frac{1}{2}\left(x+\sqrt{x^2-\lambda}\right)$:
$$-\frac{563}{1512}-\frac{43}{48}\left(y+\frac{\lambda}{y}\right)+\left(y+\frac{\lambda}{y}\right)^3=0\Longleftrightarrow$$
Substitute $\lambda=\frac{43}{144}$ and then $u=y^3$, yielding a quadratic equation in the variable $u$:
$$u^2-\frac{563u}{1512}+\frac{79507}{2985984}=0$$
Jan Eerland
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@ArchisWelankar Because otherwise you don't eliminate the quadratic term – Jan Eerland Dec 22 '15 at 12:05
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2That's called the Tschirnhaus transformation. The rule is to make the substitution $x=z-\frac{b}{3a}$. To prove this works, make that substitution in a general cubic equation and see how the quadratic term cancels. – Alice Ryhl Dec 22 '15 at 12:07
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@ Jan Eerland thats what i was asking which is answered by Kristoffer Ryhl. – Archis Welankar Dec 22 '15 at 12:27
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No you interpreted wrongly i wanted general substitution but still thanks. – Archis Welankar Dec 22 '15 at 12:40
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Here's a concise formula so anytime you come across a cubic you can easily solve it yourself. Given,
$$Ax^3+Bx^2+Cx+D=0$$
divide by $A$ to get the simpler,
$$x^3+ax^2+bx+c=0$$
then,
$$x = \frac{1}{3}\left(-a+u_1^{1/3}+\frac{a^2-3b}{u_1^{1/3}}\right)$$
where $u$ is any non-zero root of the quadratic,
$$u^2 + (2 a^3 - 9 a b + 27 c) u + (a^2 - 3 b)^3 = 0$$
P.S. This is the formula used by Mathematica though simplified for aesthetics.
Tito Piezas III
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Just to add, taking the cube root of a number will give you three solutions $u_1^{1/3} e^{2\pi i k/3}$ for $k=0,1,2$. These will yield the three roots $x_i$ of the cubic. – Tito Piezas III Jan 11 '16 at 16:09